1. RENAL BLOOD FLOW AND CLEARANCE Dr. Shaikh Mujeeb Ahmed Assistant Professor AlMaarefa College URINARY BLOCK 313

2. Objectives Describe the concept of renal plasma clearance Use the formula for measuring renal clearance Use clearance principles for inulin, creatinine etc. for determination of GFR Use PAH clearance for measuring renal blood flow Outlines the factors affecting the renal blood flow.

3. Although kidneys constitute less than 0.5% of total body mass, they receive 20-25% of resting cardiac output Left and right renal artery enters kidney Branches into segmental, interlobar, arcuate, interlobular arteries Each nephron receives one afferent arteriole Divides into glomerulus – capillary ball Reunite to form efferent arteriole (unique) Divide to form peritubular capillaries or some have vasa recta Peritubular venule, interlobar vein and renal vein exits kidney RENAL BLOOD SUPPLY

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5. Renal blood flow and Oxygen consumption Oxygen consumption per 100 g of renal tissue is 5 ml/min. which is next highest to the heart (8ml/min). Total renal blood flow is ≈ 1200/min. RBF per 100 g tissue is ≈ 400 mL/min which is disproportionately high as compared to heart which is only 80 mL/min. A-V O 2 difference is 1.5ml O 2 /100ml of blood flow. In heart it is ≈ 10 ml O 2 /100ml.

6. Renal blood flow and Oxygen consumption 90 % of RBF goes to the renal cortex, 9% goes to outer medulla and 1% to the inner medulla

7. REGULATION OF RENAL BLOOD FLOW RBF (Q) is directly proportional to the pressure gradient (ΔP) between the renal artery and the renal vein Is inversely proportional to the resistance(R) of the renal vasculature (Q) = Δ P R The major mechanism of changing Renal blood flow is by changing Afferent or Efferent Arteriolar resistance.

8. 1) SYMPATHETIC NERVES AND CIRCULATING CATACHOLAMINES Both afferent and efferent arterioles are innervated by sympathetic nerves that act via α1 receptors to cause vasoconstriction. However,since far more α1 receptors are present on Afferent arterioles, increased sympathetic stimulation will cause a decrease in both RBF & GFR.

9. 2) ANGIOTENSIN II This is a potent vasoconstrictor. However Efferent arteriole is more sensitive to Angiotensin II. Hence low levels of Angiotensin II causes increase in GFR while high levels of Angiotensin II will decrease GFR. RBF is decreased. 3) PROSTAGLANDINS PGE 2, PGI 2 are produced locally in the kidneys – cause vasodilation of both afferent & efferent arterioles. This effect is protective for renal blood flow, it modulates the vasoconstriction produce by sympathetic & angiotensin-II 4)DOPAMINE At low levels Dopamine dilates Cerebral, Cardiac, Splanchnic & Renal arterioles and constricts Skeletal Muscle and Cutaneous arterioles. Hence low dose Dopamine can be used in the treatment of hemorrhage.

10. AUTOREGULATION OF RENAL BLOOD FLOW 1.Myogenic theory 2. Tubuloglomerular feedback by Juxta Glomerular Apparatus (JGA)

11. Assessing Kidney Function Albumin excretion (microalbuminuria) Plasma concentration of waste products (e.g. BUN, creatinine) Urine specific gravity, urine concentrating ability Imaging methods (e.g. MRI, PET, arteriograms, iv pyelography, ultrasound etc) Isotope renal scans Biopsy Clearance methods (e.g. 24-hr creatinine clearance) etc

12. Clearance Renal clearance of a substance is the volume of plasma completely cleared of a substance per min by the kidneys. Clearance is a general concept that describes the rate at which substances are removed (cleared) from the plasma.

13. Clearance Technique Renal clearance of a substance is the volume of plasma completely cleared of a substance per min. Cs = Us x V Ps Cs x Ps = Us x V Where: Cs = clearance of substance S Ps = plasma conc. of substance S Us = urine conc. of substance S V = urine flow rate

14. INULIN Glucose Urea PAH

15. For a substance that is freely filtered, but not reabsorbed or secreted (inulin, 125 I-iothalamate, creatinine), renal clearance is equal to GFR Use of Clearance to Measure GFR amount filtered = amount excreted GFR x P in = U in x V GFR = P in U in x V

16. Calculate the GFR from the following data: P inulin = 1.0 mg / 100ml U inulin = 125 mg/100 ml Urine flow rate = 1.0 ml/min GFR = 125 x 1.0 1.0 = 125 ml/min GFR = C inulin = P in U in x V

17. Clinically it is not convenient to use inline clearance – to maintain a constant plasma concentration it must be infused continuously throughout measurement Creatinine clearance is used as a rough estimate of GFR Normal plasma serum creatinine level is 1mg/100ml.

18. Effect of reducing GFR by 50 % on serum creatinine concentration and creatinine excretion rate Figure 27-18; Guyton and Hall

19. Steady-state relationship between GFR and serum creatinine concentration Figure 27-19; Guyton and Hall

20. Sample problem In a 24hr period, 1.44 L of urine is collected from a man receiving an infusion of inulin. In his urine, the [inulin] is 150mg/ml, and [Na+] is 200 mEq/L. In his plasma, the [inulin] is 1mg/mL, and the [Na+] is 140mEq/L What is the clearance ratio for Na+, and what is the significance of its value?

21. Theoretically, if a substance is completely cleared from the plasma, its clearance rate would equal renal plasma flow. Use of Clearance to Estimate Renal Plasma Flow Cx = renal plasma flow

22. Paraminohippuric acid (PAH) is freely filtered and secreted and is almost completely cleared from the renal plasma Use of PAH Clearance to Estimate Renal Plasma Flow 1. amount enter kidney = RPF x P PAH 3. ERPF x P pah = U PAH x V ERPF = U PAH x V P PAH ERPF = Clearance PAH 2. amount entered = amount excreted ~ ~ 10 % PAH remains Copyright © 2006 by Elsevier, Inc.

23. PAH clearance example

24. Calculate the RPF from the following data: P PAH = 0.01 mg / 100ml U PAH = 5.85 mg/100 ml Urine flow rate = 1.0 ml/min ERPF = 5.85 x 1.0 0.01 = 585 ml/min ERPF = C PAH = P PAH U PAH x V

25. To Calculate Actual RPF, One Must Correct for Incomplete Extraction of PAH E PAH = A PAH - V PAH A PAH RPF = ERPF E PAH normally, E PAH = 0.9 i.e PAH is 90 % extracted A PAH =1.0 V PAH = 0.1 = 1.0 – 0.1 1.0 = 0.9

26. To Calculate Actual RPF, One Must Correct for Incomplete Extraction of PAH E PAH = A PAH - V PAH A PAH normally, E PAH = 0.9 i.e PAH is 90 % extracted A PAH =1.0 V PAH = 0.1 = 1.0 – 0.1 1.0 = 0.9 RPF = 585 0.9 = 650

27. RBF = RPF 1 - Hct RBF = 650 0.55 = 1182 RBF = 650 1 - 0.45

28. Calculation of Tubular Reabsorption Reabsorption = Filtration -Excretion Filt s = GFR x Ps Excret s = Us x V

29. Calculation of Tubular Secretion Secretion = Excretion - Filtration Filt s = GFR x Ps Excret s = Us x V

30. Clearances of Different Substances Clearance of inulin (C in ) = GFR if Cx < Cin: indicates reabsorption of x Clearance of PAH (C pah ) ~ effective renal plasma flow SubstanceClearance (ml/min inulin125 PAH600 glucose 0 sodium 0.9 urea 70 Clearance creatinine(C creat ) ~ 140 (used to estimate GFR) if Cx > Cin: indicates secretion of x

31. References Human physiology by Lauralee Sherwood, seventh edition Text book of physiology by Linda.s contanzo,third edition Text book physiology by Guyton &Hall,11 th edition